I am reading the Balazard Saias and Yor paper where he defines,
$s=s(z)=\frac{1}{1-z}$ and $$f(z)=(s-1)\zeta(s)$$ where $\zeta$ is the Riemann zeta function.Then by Jensen's formula we have for $f(0)\neq 0$ and $r<1$,
$$\frac{1}{2\pi}\int_{-\pi}^{\pi}log\ |f(re^{i\theta})|d\theta=log\ |f(0)|+\sum_{|\alpha|<r,f(\alpha)=0} log \ \frac{r}{|\alpha|}$$
For applying Jensen's formula we need that $f(z)$ has no zeros on $|z|=r$.
Question Show that $f(z)$ has no zeros on $|z|=r$, $r<1$.
Attempt
$$f(z)=(s-1)\zeta(s)$$
$$f(z)= \frac{z}{1-z}\zeta(\frac{1}{1-z})$$
Since $|z|=r$ so write $z=re^{i\theta}$
$$\zeta(\frac{1}{1-z})= \zeta(\frac{1}{1-re^{i\theta}}) $$
$$\zeta(\frac{1}{1-z})= \zeta(\frac{1}{1-rcos\theta-irsin\theta}) $$
$$\zeta(\frac{1}{1-z})= \zeta(\frac{1-rcos\theta+irsin\theta}{(1-rcos\theta)^2+r^2sin^2\theta}) $$
$$\zeta(\frac{1}{1-z})= \zeta(\frac{1-rcos\theta+irsin\theta}{r^2-2rcos\theta+1}) $$
How to show that $f(re^{i\theta})\neq 0$ on $|z|=r, r<1$?.
Best Answer
Jensen's theorem is valid even if $f$ has zeroes on the circle $|z|=r$ (in practice since such circles are discrete, one avoids them easily in the usual applications of Jensen which is to count zeros in open discs)
The proof is standard assuming the result first without zeroes; then the number of zeroes on the circle $|z|=r$ is finite ($f$ is analytic in a neighborhood of $|z|=r$ in Jensen), hence let them be $a_1,..a_n, |a_k|=r$ counted with multiplicity.
Hence $f=(z-a_1)..(z-a_n)g$ with $g$ same zeroes as $f$ except those on the circle and $\log |f|=\log |g|+\log |z-a_1|+...\log |z-a_k|$
But now applying Jensen for $g$, using the easy to prove fact that $\int_0^{2\pi}\log |r(e^{i\theta}-e^{i\theta_0})|d\theta=2\pi \log r$, the fact that $|\log f(0)|=n\log r+|\log g(0)|$ and the observation that $\log r/|a_k|=0$ gives the result.